Question

A planet moves around the Sun in an elliptical orbit (see Fig. 8.39 ). (a) Show that the external torque acting on the planet about an axis through the Sun is zero. (b) since the torque is zero, the planet's angular momentum is constant. Write an expression for the planet's angular momentum in terms of its mass \(m,\) its distance \(r\) from the Sun, and its angular velocity \(\omega\) (c) Given \(r\) and a, how much area is swept out during a short time At? [Hint: Think of the area as a fraction of the area of a circle, like a slice of pie; if \(\Delta t\) is short enough, the radius of the orbit during that time is nearly constant.] (d) Show that the area swept out per unit time is constant. You have just proved Kepler's second law!

Short answer

Question: Show that the external torque acting on a planet in an elliptical orbit around the sun is zero, and determine the planet's angular momentum in terms of its mass (m), distance (r) from the sun, and its angular velocity (ω). Answer: The external torque acting on the planet is zero because the gravitational force exerted by the Sun is directed along the line connecting the planet and the Sun, which is the same as the direction of the planet's position vector. Since the torque vector is perpendicular to both the position vector and force vector, their cross product will be zero. The planet's angular momentum can be expressed as L = m r^2 ω.

Step-by-step solution

(a) External torque calculation

To find the external torque acting on the planet, we have to consider the gravitational force \(F_g\) that the Sun exerts on the planet. Using \(\tau = \vec{r} \times \vec{F}\), where \(\vec{r}\) is the position vector of the planet, and \(\vec{F}\) is the force vector. In this case, the force \(\vec{F}\) is directed along the line connecting the planet and the Sun, which is the same as the direction of \(\vec{r}\). Since the torque vector is perpendicular to both position vector and force vector, their cross product will be zero. Therefore, \(\tau = 0\).

(b) Angular momentum expression

Angular momentum \(\vec{L}\) can be expressed in terms of linear momentum \(\vec{p}\) and position vector \(\vec{r}\) as \(\vec{L} = \vec{r} \times \vec{p}\). In polar coordinates, with radius \(r\) and angular velocity \(\omega\), the linear velocity of the planet can be expressed as \(\vec{v} = r \omega \hat{\theta}\). We also know \(\vec{p} = m\vec{v}\). Plugging these values into the equation for angular momentum, we get \(\vec{L} = \vec{r} \times m(r \omega \hat{\theta})\). We can simplify this to \(\vec{L} = m r^2 \omega \hat{k}\).

(c) Area swept during a short time

In a short period, \(\Delta t\), the planet sweeps a small area, which can be considered as a sector of a circle with radius \(r\) and angle \(\Delta \theta = r \omega \Delta t\). The area of this sector can be expressed as \(\Delta A = \frac{1}{2} r^2 \Delta \theta\). Plugging in the value of \(\Delta \theta\), we get \(\Delta A = \frac{1}{2} r^2 r \omega \Delta t\). Simplifying, we get \(\Delta A = \frac{1}{2} r^3 \omega \Delta t\).

(d) Area swept per unit time is constant

We know that since the external torque is zero, the angular momentum is constant. So, \(L = m r^2 \omega=k\) where k is a constant. Solving for \(\omega\), we get \(\omega = \frac{k}{m r^2}\). Now, to find the area swept per unit time, we need to divide the area swept by the time interval: \(\frac{\Delta A}{\Delta t} = \frac{1}{2} r^3 \omega\). Replacing \(\omega\) with the expression we found above, we get \(\frac{\Delta A}{\Delta t} = \frac{1}{2} r^3 \frac{k}{m r^2}\). Simplifying, we see that the area swept per unit time is constant and equal to \(\frac{k}{2m}\). This proves Kepler's second law, which states the rate of change of the area swept by a planet in its orbit is constant.